Total differentiability at the origin when $f_x$ is continuous

Substituting in $\vector x = \vector 0$, being differentiable at zero requires that $\exists D$ such that
$$
f(\vector h) - f(\vector 0) = D(\vector 0)\vector h + \varepsilon(\vector h), ~\frac{\varepsilon(\vector h)}{\norm{\vector h}} \to 0, 
\\ \text{ or } \\ 
\lim_{\vector h \to \vector 0} \abs{\frac{f(\vector h) - f(\vector 0) - D(\vector 0)\vector h}{\norm{\vector h}}} = 0.
$$
We'll make use the mean value theorem. The general statement is
$$
f' \text{ exists and continuous on } (a,b)
\implies \exists A \in [a,b] \suchthat \\
f(b) - f(a) = f'(c)(b-a)
$$
and here we'll use
$$
f_x \text{ exists and continuous on } (0, t) 
\implies \exists A \in [0, t] \suchthat \\ 
f(t, y) - f(0, y) = f_x(A, y)(t - 0) = tf_x(A, y) \\
\implies \color{purple} f(t, y) - f(0, y) = tf_x(A, y)
$$
where moreover 
$$
\lim_{t\to 0} f_x(A, y) = f_x(\lim_{t\to 0} A, y) = f(0, y)
$$ 
since $f_x$ was assumed to be continuous.

We'll also use the "best linear approximation" definition of differentiability. In general, it states
$$
f' \text{ exists on } (a,b) \\
\implies f(x) = f(a) + f'(a)(x-a) + o(x-a),~ \frac{o(x-a)}{x-a} \to 0 \\
\implies f(x) - f(a) = f'(a)(x-a) + o(x-a),~ \frac{o(x-a)}{x-a} \to 0 \\
\implies f(b) - f(a) = f'(a)(b-a) + o(b-a),~ \frac{o(b-a)}{b-a} \to 0
$$
and here we'll take $a=0, b=t$, so $x-a = x$ and $b-a = t$, and use the variant
$$
f_y \text{ exists on } (0, t) \\ 
\implies f(x, t) - f(x, 0) = f_y(x, 0)(t - 0) + o(t),~ \frac{o(t)}{t} \to 0 \\
\implies \color{green}f(x, t) - f(x, 0) = tf_y(x,0) + o(t), ~\frac{o(t)}{t} \to 0.
$$
We can then write
$$\begin{align*}
f(h_1, h_2) - f(0, 0)
&= {\color{green}f(h_1, h_2) - f(h_1, 0)} + {\color{purple}f(h_1, 0) - f(0, 0)} \\
&= {\color{green}h_2f_y(0, 0) + o(h_2)} + {\color{purple}h_1f_x(A, 0)}\\
&= (h_1 f_x(0, 0) - {\color{red}h_1 f_x(0, 0)}) + {\color{purple}h_1f_x(A, 0)} + {\color{green}h_2f_y(0, 0) + o(h_2)} \\
&= h_1 f_x(0, 0) + {\color{green}h_2 f_y(0, 0)} + h_1({\color{purple}f_x(A,0)} - {\color{red}f_x(0,0)}) + {\color{green}o(h_2)}
\end{align*}$$

Now, since $f_x$ is continuous, ${\color{purple}f_x(A,0)} - {\color{red}f_x(0,0)} \to 0$ and so $h_1(f_x(A,0) - f_x(0,0)) = o(h_1)$. We thus have
$$\begin{align*}
f(h_1, h_2) - f(0, 0)
&= h_1 f_x(0, 0) + h_2 f_y(0, 0) + o(h_1) + o(h_2) \\
&= h_1 f_x(0, 0) + h_2 f_y(0, 0) + o(\norm{\vector h}) \\ \\
&\implies 
f(h_1, h_2) - f(0, 0) - h_1 f_x(0, 0) - h_2 f_y(0, 0) = o(\norm{\vector h})
\end{align*}$$

and so if we define 
$$D(\vector h) = h_1 f_x(0, 0) + h_2 f_y(0, 0),
$$ 

we can take absolute values to find that 
$$
\abs{f(\vector h) - f(\vector 0) - D(\vector h)} = o(\norm{\vector h}) \\
\text{ and so } \\
\lim_{\vector h \to \vector 0} \abs{\frac{f(\vector h) - f(\vector 0) - D(\vector h)}{\norm{\vector h}}} = 0
$$

which forces $f$ to be totally differentiable at $\vector 0$ with derivative $D$.

Trivial algebraic identity by adding and subtracting \(f(0, h_2)\).

For fixed \(h_2\), the function \(g(t) = f(t, h_2)\) is differentiable on the interval between \(0\) and \(h_1\) because \(f_x\) exists in a neighborhood of \((0,0)\). By the single-variable MVT, \(g(h_1) - g(0) = h_1 g'(\theta h_1) = h_1 f_x(\theta h_1, h_2)\).

By definition of the partial derivative \(f_y(0,0) = \lim_{h_2 \to 0} \frac{f(0, h_2) - f(0,0)}{h_2}\).

Substitute 3 and 4 into 2 and apply the triangle inequality: \[f(\mathbf{h}) - f(\mathbf{0}) - h_1 f_x(\mathbf{0}) - h_2 f_y(\mathbf{0}) = h_1 [f_x(\theta h_1, h_2) - f_x(0,0)] + r(h_2).\]